Question

Calculate the solubility (in m units) of ammonia gas in water at 298 k and a partial pressure of 4.50 bar . the henry's law constant for ammonia gas at 298 k is 58.0 m/atm and 1 bar=0.9869 atm.

Answer 1

The solubility of ammonia gas in water at 298 K and a partial pressure of 4.50 bar is calculated using Henry's Law, resulting in a solubility of 257.58 M after converting the pressure to atmospheres.

Step-by-step explanation:

To calculate the solubility of ammonia gas in water at 298 K and a partial pressure of 4.50 bar, we use Henry's Law, which states that the solubility of a gas in a liquid is directly proportional to the partial pressure of the gas above the liquid. Given that the Henry's Law constant for ammonia at 298 K is 58.0 m/atm and 1 bar is equivalent to 0.9869 atm, we first convert the pressure in bars to atmospheres:

Pressure in atm = 4.50 bar × 0.9869 atm/bar = 4.44105 atm.

Next, apply Henry's Law to find the solubility in molarity (m units):

Solubility (M) = Henry's Law constant (K) × Partial pressure (P)

Solubility (M) = 58.0 M/atm × 4.44105 atm = 257.5809 M.

Therefore, the solubility of ammonia gas in water at 298 K and a partial pressure of 4.50 bar is 257.58 M.

Answer 2

We'll use the formula Solubility = K * P where K is the Henry's constant and P is the pressure in atm.
If 1 bar = 0.9869 atm.
Then 4.5 bar = X atm
X = 4.4410
So Solubility = 58 * 4.4410 = 257.574 M